-1 g 3.i8 - solvents and solvent effects r nh gthe solvent molecules complex to the copper metal...
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IIIB Paper 3 May 2011 answers
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3.I8 - Solvents and Solvent Effects
a) i)
NH4+(g) + NMe3(g) NH3(g) + HNMe 3
+(g)
Gr0 = -92 kJ mol -1
NH4+(aq) + NMe 3(aq) NH3(aq) + HNMe 3
+(aq)
Gr0 = -3 kJ mol -1
Gsolv0
= -331 kJ mol -1Gsolv
0
= -242 kJ mol -1
(1 mark each for labelling ΔGrxn for the gas phase, ΔGrxn in water, ΔGsolv in the
gas phase and ΔGsolv in water – numerical values are unnecessary)
ii) ΔGrxn = -3 + (-331 + 242) = -92 kJ/mol. (1 mark for value, 1 mark for units)
iii) N(CH3)3 is the stronger base in the gas phase (1 mark) because the ΔGrxn is
negative (1 mark). Also acceptable is a legitimate chemical reason, such as alkyl
groups donating electron density into the nitrogen‟s lone pair or the increased size
giving increased polarizability.
iv) There is very little difference in the solvation of the neutral molecules (1 mark)
since the lone pair is accessible. However, the alkyl groups prevent solvation of the
ammonium centre in the charged molecules (1 mark), reducing hydration and
therefore lowering the stability of the tetramethylammonium centre.
N+
H
H
H
H
HH
O
O O
O
H
H
HH
H
H
N+
CH3
H
CH3
CH3 O
H
H
(Drawings optional)
v)
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(0.5 marks each for the shared proton, the partial bonds, the delocalized charge
and the stereochemistry)
vi)
(0.5 marks for having each state in approximately the correct location according
to the known sizes of the ΔG’s, 0.5 marks for labelling each ΔG)
vii) The solvent effect on the rate of reaction will be setermined by the relative
solvation of the reactants and the transition state. (0.5 marks each)
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b) i) A solvatochromic scale could be developed by using νmax as an energy parameter.
(1 mark)
ii)
(1 mark for square planar Cu(acac)2, 1 mark for octahedral solvated complex)
The solvent molecules complex to the copper metal centre (1 mark) through electron
pair donation (Lewis base) (1 mark).
iii) The energy of transition will be lower in more polar solvents because the solvent
donates electron density to the metal centre (0.5 marks) thus lowering the energy gap
between the ground state and the excited state (0.5 marks).
c)
i) An LSER will require an energy term (ln (k2) and a parameter (the Kirkwood
parameter). Calculating these:
Solvent ε
k2 (s-1
M-1
) ln (k2)
Toluene 2.3 0.232 1.0*10-4
-9.21
Dichloromethane 8.4 0.416 1.0*10-2
-4.61
Acetone 21.5 0.466 3.5*10-2
-3.35
Nitrobenzene 37.3 0.480 5.0*10-2
-3.00
Fitting a line to the first and fourth points (any 2 will suffice):
Slope = (-3.00+9.21)/(0.480-0.232) = 25
Intercept = (-3.00) – (25)*(0.480) = -15
So the LSER would be:
ln (k2) = 25 * – 15
(1 mark for slope, 1 mark for intercept, 1 mark for using ln k2)
ii) ε = 5.4, so = 0.373
Plugging this in to the LSER yields:
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ln (k2) = (25) * (0.373) – 15 = -5.675
k2 = exp (-5.675) = 3.4*10-3
s-1
M-1
(0.5 marks for value, 0.5 marks for units)
iii) The rate in methanol from the LSER:
iv) ε = 32, so = 0.477
ln (k2) = (25) * (0.477) – 15 = -3.08
k2 = exp (-3.08) = 4.6*10-2
s-1
M-1
Which is much larger than the experimental value. (1 mark)
Methanol is a hydrogen bond donating solvent, and none of the ones used in
the LSER were, so hydrogen bonding must be causing the problem. (0.5
marks)
Methanol can hydrogen bond to the amine, thus lowering the energy of the
reactants relative to the transition state, which will increase ΔG‡ and slow the
reaction down. (0.5 marks)
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3. O5 REACTIVE INTERMEDIATES
Answer ALL parts, (a)-(c)
(a) Suggest a radical-based method for carrying out ONE of the following transformations,
giving reagents and a mechanism. More than one step may be required.
(5 marks)
(b) Provide a mechanism for the following transformation commenting on any aspects of
selectivity.
(5 marks)
CONTINUED ON NEXT PAGE
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(c) Give the products and mechanisms for THREE of the following reactions, noting
carefully important regiochemical or stereochemical aspects.
(5 marks each)
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O10. ORGANIC MOLECULAR MODELLING. 2010-2011 Outline Answers.
a)
i) A reasonable discussion on these principle aspects is expected (but not a comprehensive one): Molecular
Mechanics or Quantum mechanics provides a total energy which is expanded as a Taylor theorem to
-G.H-1
. The
technique of geometry optimisation uses this algorithm to minimise the energy for all 3N-6 degrees of
freedom, using one of three approximate and one exact way of specifying the H matrix.
ii) Mechanics should not be used, since the parameters for a sp2 carbon (carbene) and those for sp2 carbon
(double bond) cannot be compared. Also, mechanics cannot be used to break or form bonds. Quantum
mechanics is the only method that can be used, and this can provide a free energy for the reaction, and via
a reaction coordinate calculation, a free energy of activation for the reaction.
iii) Visualisation: Con, no coordinates available since it has never been made. Molecular mechanics: Pro; lots
of close H…H contacts expected, which are well handled via the non-bonded (van der Waals) terms. Con;
the C=C is likely to be much longer than normal. Mechanics is parameterised on normal bond lengths,
and the assumption of a quadratic potential for the stretching of a bond, so the longer the bond, and the
less accurate its prediction. Also cross terms (e.g. stretch-bend) are not included in most implementations,
such as MM2 (i.e. a long length may impact the angles at the affected carbon). DFT: Pros; Good at
forming/breaking bonds, good at dealing with unusual bond types, good at computing the entropic
contributions.
b)
i) Mechanics would give a good analysis of the dispersion interactions between the two monomers, and a
less good analysis of the unusual hydrogen bonding present. It would not routinely enable the entropy of
dimerization to be calculated. DFT methods would need to include a dispersion-correction to compete
with mechanics, but can be made to include the entropic contribution.
ii) Visualisation (X-ray structure) shows the hydrogen bonding involving the OH group to be unusually to a
-face, which precludes mechanics (unlikely to be parameterised for this kind of H-bond). DFT can
provide information about the electronic distribution in the region of the H-bond. Evaluating that
distribution by inspecting the molecular orbitals does not allow the inductive effect of the CF3 group to be
interaction nicely. Another suitable method is quantum topology, involving calculating the dimer, and
locating intermolecular (bond) critical points (BCPs) in the electronic distribution.
iii) Visualisation can provide distances between X or Y and close atoms; the strengths could be very
approximately estimated from comparing these distances with the sum of the van der Waals radii of those
atoms. Anther method would be the quantum topological method which can be used to identify any
intramolecular bond critical points (BCP) in the electron density distribution of the dimer, found from X
or Y to any other atoms they are not already covalently bonded to. The value of the electron density at the
BCP is approximately linearly related to the strength of the interaction.
c)
i) This is a reaction, involving both forming and breaking bonds. So a density function (DFT or semi-
empirical) method is appropriate. A rate constant is obtained from the free energy of activation for the
reaction, so a transition state for the process must be located. This is done by the process of geometry
minimisation, but using an algorithm that ensures the 2nd
derivative (Hessian) matrix has exactly one
negative root (eigenvalue) and that the resulting (eigen)vectors (displacement coordinates) correspond to
the desired reaction. The transition state can have both twist-boat and chair conformations which both
have to be located.
ii) The factors are predominantly steric interactions (or attractions) between the two methyl groups and
X=Me. The transition state for the reaction in either a chair or boat conformation has to be modelled with
a method capable of breaking/forming bonds (DFT or semi-empirical), and ideally of handling any van
der Waals interactions (by including a dispersion correction) and the free energy of that transition state is
compared for the two diastereomeric forms to establish their relative concentrations.
iii) X has to be chiral (a chiral auxiliary. The same procedure as in the previous part is followed, but now all
four (in equivalent because of chiral X) transition states have to be located and their relative free energies
compared.
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3.P6 - Molecular Electronic Materials
Answer part a) AND EITHER part b) OR part c)
a) Answer ALL parts of this question.
i) Answer should include discussion of Frenkel excitons. High binding energy of
excitons in organic semiconductors means that at room temperature very few free
electrons and holes. To overcome this problem use an interface or electron donor-
acceptor interface to overcome columbic attraction of electrons and holes. Draw
labelled diagram of donor-acceptor interface (3 marks)
ii) 3 marks: Absorption profile should follow:
With k = 0.003 / angstrom then absorption profile should look like
)exp( kxII o
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0 500 1000 1500 2000
0.0
0.2
0.4
0.6
0.8
1.0
I/I o
Thickness of polymer layer / Angstroms
semiconductor X semiconductor Y
heterojunction
iii) Define n as Ndis/Ncreated then:
Assumption: kL << 1 Therefore: exp(+kL) approximately = 1 + kL
exp(-kL) approximately = 1 - kL
𝑁𝑑𝑖𝑠𝑁𝑐𝑟𝑒𝑎𝑡𝑒𝑑
= 𝐼𝑜 exp −𝑘𝑥 𝑑𝑥𝑥𝑖+𝐿
𝑥𝑖−𝐿
𝐼𝑜 exp −𝑘𝑥 𝑑𝑥∞
0
𝑁𝑑𝑖𝑠𝑁𝑐𝑟𝑒𝑎𝑡𝑒𝑑
=
−𝐼𝑜𝑘
exp −𝑘𝑥 𝑥𝑖−𝐿
𝑥𝑖+𝐿
−𝐼𝑜𝑘
exp −𝑘𝑥 0
∞
𝑁𝑑𝑖𝑠𝑁𝑐𝑟𝑒𝑎𝑡𝑒𝑑
=exp −𝑘𝑥𝑖 − 𝑘𝐿 − exp −𝑘𝑥𝑖 + 𝑘𝐿
−1
𝑁𝑑𝑖𝑠𝑁𝑐𝑟𝑒𝑎𝑡𝑒𝑑
=exp −𝑘𝑥𝑖 1 − kL − 1 − kL
−1
𝑁𝑑𝑖𝑠𝑁𝑐𝑟𝑒𝑎𝑡𝑒𝑑
= 2𝑘𝐿 𝑒𝑥𝑝(−𝑘𝑥𝑖)
𝑁𝑑𝑖𝑠𝑁𝑐𝑟𝑒𝑎𝑡𝑒𝑑
=exp −𝑘𝑥𝑖 exp −𝑘𝐿 − exp(+𝑘𝐿)
−1
(5 marks)
iv) Using the values of k = 0.003 / Angstrom and L = 60 angstroms, thickness of
polymer X = 1000 Angstroms then n = 0.017. This would significantly reduce QE.
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Can overcome this problem by using a blend of the two polymers X and Y where
the interfaces are structured on exciton diffusion length of 60 Angstroms.
(4 marks)
b) Answer ALL parts of this question.
i) 3 marks: Charge transport in an organic semiconductor is a thermally activated
hopping process. Each hop assisted by an electric field. In practice carrier
transport proceeds via a series of hops between molecules. Such hops are the rate
limiting step in transport. In practice the smaller the interchain (inter molecule)
distance, the better the charge transport
Each hop / jump can be assisted by an applied field, thus there is a field-
dependence to the mobility (termed a Poole-Frenkel type mobility). Mobility can
be given by:
ii) 2 marks: An organic solar cell: need to high electron and hole mobility in the
electron transporting and hole transporting material respectively. High charge
carrier mobility needed to minimize charge recombination between the
photogenerated electrons and holes. The QE depends on the probability of
electrons and holes not recombining.
2 marks: for an OLED need to balanced electron and hole mobility to avoid
leakage currents. Easier achieved in multilayered structures as opposed to single
layer devices.
iii) 2 marks: Mobility = 5.36 x 10 -6
m2V
-1s
-1
c) Answer ALL parts of this question.
i) 2 marks: Fowler Nordheim: A particular case of quantum mechanical tunnelling
through a barrier. If there is an injection „barrier‟ for electron or hole injection,
applying a voltage will cause a field drop across insulator. This creates a barrier
across which electrons or holes can tunnel.
𝐽 ∝ 𝐸2𝑒𝑥𝑝 −𝐵𝜙3 2
𝐸
k = Bf3/2
E = electric fieldJ is the currentf = barrier height (same as fb)
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• 2 marks: Thermionic emission: Here, electrons or holes have enough thermal energy
to „hop‟ over the barrier into the organic semiconductor. However barrier height is
actually lowered due to an „image-force‟. The image force results from the attractive
Coulombic force that any charge experiences close to a metallic surface. Here, the
barrier is lowered by an amount phi(f)
ii) Answer should include:
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ff )(ext
External quantumyield
Depends on transport of electron and hole polarons (i.e how easy electron and hole polarons move across film)
Light out couplingefficiency. Effected byself-absorption and refractive index of material ( ~ 1/2n2)
Intrinsic property of organic material. Intrinsic quantum efficiency for radiativedecay
f = tf / tr
Fraction of total excitons formed which can result in radiative decay. Intrinsic property of organic material
Assuming refractiveIndex losses only
Can optimize gamma by chosen multilayered structured to reduce leakage currents. In
fluorescent materials alpha = 0.25 (due to singlet/triplet ratio). Can optimize alpha by using
triplet emitters – can use compounds with heavy metals. Can optimize phi by using materials
with high QE of luminescence.
Light out coupling can be improved by controlling refractive index of material.
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3.P10 - Soft Condensed Matter
Answer part (a) and EITHER part (b) OR part (c).
a) Answer ALL parts of this question
i) With the aid of suitable graphs briefly discuss how the range of the interactions
influences the coexistence diagram of soft materials.
(3 marks)
ANSWER
When the range is long enough, we have got a stable liquid phase, for short-range interactions
the liquid phase is not stable and metastable fluid-fluid and solid-solid phases might appear.
The mestastable phases appear because the relaxation to the thermodynamic stable phases is
very long, these mestastable phases can be observed in real situations.
Diagrams (2 marks), explanation (1 mark).
(3 marks)
ii) The Bjerrum length is an estimate of the interaction range between two charged
colloids in a suspension. In an experiment, it is observed that the colloid liquid phase
becomes unstable when the interaction range is about 20 nm. Estimate the dielectric
constant of the suspension.
Data [Colloids charge = 5 e, T = 300 K]
ANSWER
The Bjerrum length is defined by,
If the liquid phase becomes unstable at 2 nm, this means that the interaction is of the order of
kT at that distance. Hence, substituting in the Bjerrum length equation we get
lB qiq je
2
4okBT
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lB qiq je
2
40kBT
202e2
lB 40kBT
52e2(C2)
2108(m)4 (8.8541012J1C2m1)kB (J /K)300(K) 69.6
(3.5 marks)
Correct equation for Bjerrum length (1 mark), method/justification (1 mark), correct result
(1.5 mark)
iii) For a glass, sketch the dependence of the entropy with temperature, indicating the
location of the Kauzmann temperature. Briefly explain the physical meaning of
this temperature.
How does the speed of undercooling affect the entropy dependence with
temperature? Justify your answer.
(3.5 marks)
ANSWER
The diagram should look like.
The Kauzmann temperature corresponds to the intersect of the entropy of the supercooled
liquid with the entropy curve of the crystalline solid. Because the entropy of the glass cannot
be smaller than that of the crystal the Kauzmann T marks a lower limit in the experimental
glass transition temperature.
The slower the liquid is cooled the longer the time available for configurational sampling and
hence the colder the liquid can become before falling in the glass state. Hence the glass
transition temperature decreases with the speed of undercooling.
Diagram (1.5 marks), Kauzmann temperature (1 mark), dependence of S with undercooling
(1 mark).
(3.5 marks)
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iv) The relaxation time for configuration rearrangements in a glass is given by the
Vogel-Fulcher equation,
t t 0 expB
T T0
The glass transition temperature of an experiment performed in a 104
seconds
timescale is 200 K. Estimate the glass transition temperature when the experimental
time increases by two orders of magnitude, 106
seconds.
Comment on your result.
[Data: B = 103
K, T0=50 K]
(5 marks)
ANSWER
We identify the experimental time with
t . Hence,
lnt1 lnt 0
B
T1 T0
lnt 2 lnt 0 B
T2 T0
lnt 2
t1
B
T2 T0
B
T1 T0
which simplifies to:
1
Blnt 2
t1
1
T1 T0
1
T2 T0
hence we get, T2 = 138.7 K
The glass transition temperature decreases; this is consistent with a slower cooling rate in the
experiment, i.e., with a longer experimental time. This gives more time to the molecules to
relax towards the equilibrium positions (the equilibrium positions are not reached in the
experimental time).
Correct identification of tau with experimental time (1 mark) , correct derivation of equation
and result (3 marks), justification (1 mark).
b) Answer ALL parts of this question
i) For a homogeneous nucleation process, sketch the dependence of the Gibbs
free energy with the radius of a spherical nucleus. Justify your answer by
discussing the different contributions to the Gibbs free energy.
(3 marks)
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ANSWER
The Gibbs free changes as
Two terms contribute to this. A volume term (blue line), ~r^3, which takes into
account the gain in free energy associated to the formation of the crystal, and a
surface term that is connected to the work required to create the crystal-liquid
interface (red line), ~r^2. The balance of these contribution results in a maximum,
this is the activation free energy for the formation of the critical nucleus of radius
r*.
Correct sketch (1.5 marks), explanation, (1.5 marks). Total 3 marks
ii) Show that the radius of the critical nucleus for homogeneous nucleation is
given by the following equation:
Define all the terms in this equation.
(7 marks)
ANSWER
-gamma_sl is the solid-liquid surface tension
-Tm is the melting temperature
-Delta H_m: is the enthalpy of melting
-Delta T: is the undercooling, = Tm – T, T is the temperature of homogeneous nucleation.
Derivation.
Start with the definition of the Gibbs free energy
G(r) Gb4 /3r3 sl4r2
The critical nucleus is defined by the radius that maximizes the free energy, i.e.,
r* 2 sl TmHmT
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G
r
r r*
0, hence
Gb4r2 sl8r 0 r[Gb4r sl8]0
Rearranging we get,
r* 2 slGb
(1)
Now we recall the definition of entropy in terms of the Gibbs free energy.
Sm GlT
P
GsT
P
Hm
Tm at equilibrium
Considering,
S (G /T)P and that the derivative is constant for the undercooling range
T Tm T , we get the new equation.
Gb Hm
TmT and substituting in (1)
r* 2 slHm
Tm
Tas requested.
Correct definitions (1 mark), equation for the free energy (1 mark), stationary condition and
critical radius --equation 1 above-- (2 marks) connection between Delta G_b and Delta H_m
and final result ( 3 marks).
c) Answer ALL parts of this question
i) A colloidal suspension coagulates when the average distance between the colloids
is 1 nm. Estimate the molar concentration of KCl needed to destabilize the
suspension at 300 K.
[Data: water dielectric constant: 78]
(5 marks)
ANSWER
According to the DLVO theory the coagulation will occur when
Dh0 ~ 1 where kappa_D is
the Debye length and h_0 is the separation between the colloids. Hence
D
e2 izi2
i1
0kBT
1/ 2
using the condition from the DLVO theory:
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(Dh0)20kBT
2e2z2 KClh0
2
1 78 8.8541012(J1C2m1) kB (JK1) 300(K)
2 (1.6021019)2(C2) 1
KCl 5.5 107 /h0
2 5.5 1025
ion pairs/m^3. In mols 0.09 M.
(1 mark) recognize kappa_d h_0 ~ 1, (1 mark) state Debye length equation, (3 marks) correct
calculation of the concentration in right units M.
ii) Experimental studies have shown that the van der Waals interactions between
colloidal particles dispersed in a solvent can be repulsive, attractive or zero
depending on the suspension composition.
Explain these observations. You answer should contain a reference to the relevant
equations.
(5 marks)
ANSWER
The van der Waals interactions between colloidal particles can be written as:
Where h is the surface to surface distance, and R1 and R2 the radii of the colloids.
A is the Hamaker constant which depends on the medium and colloid
compositions through
Where rho_ci and alpha_ci are the density and polarizability of the colloids and
rho_s and alpha_s the density and polarizability of the solvent. Hence for colloids
of different composition if one of the terms in the brackets is positive and the
other is negative A<0 and the van der Waals interaction would be repulsive. For
colloids of the same composition, A>0 attractive as
A(cc ss)2 or “zero”
when
cc ss. The latter situation can be achieved by using “refractive index
matching”,as shown by the Clausius Mosotti equation.
(2 marks),van der waals equation in terms of Hamaker constant and Hamaker
constant equation in terms of alpha and rho (2 marks) explanation
repulsive/attractive interactions (1 mark) explanation zero interactions (
clausius mosotti equations).
Uvdw (h) A
6h
(R1R2)
(R1 R2)
A(c1c1 ss)(c2c2 ss)
n2 1
n2 2 4 /3